Solving time-independent Schrödinger equation

\[ \begin{equation} \hat{H} \Psi(x) = \left[\frac{\hat{p}^2}{2m} + V(x)\right] \Psi(x) = E\Psi(x), \end{equation} \]

i.e. finding \(\Psi(x)\) as the eigenfunctions of Hamiltonian \(\hat{H}\).

In the position representation, the momentum operator \(\hat{p} = -i\hbar \mathrm{d}/\mathrm{d}x\), so we end up with an ODE:

\[ \begin{equation} \left[-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2} + V(x)- E \right] \Psi(x) = 0. \end{equation} \]

Potential step

In piecewise-continuous potentials

\[\begin{equation} V(x) = \begin{cases} V_1,& \text{if } x< 0\\ V_2,& \text{if } x \geq 0, \end{cases} \end{equation} \]

In each domain \(i\)

\[ \begin{equation} \Bigg[\frac{\mathrm{d}^2}{\mathrm{d}x^2} + \underbrace{\frac{2m}{\hbar^2} (E - V_i)}_{k_i^2} \Bigg] \Psi_i(x) = 0 \end{equation} \]

If \(E>V_i\), we get \(k_i \in \mathbb{R}\).

General solutions with unknown amplitudes \(A_1,B_1,A_2,B_2\):

\[ \begin{align}\label{eq:solution1} \Psi_1(x) = \underbrace{A_1 e^{i k_1 x}}_{\rightarrow} + \underbrace{B_1 e^{-i k_1 x}}_{\leftarrow}, \quad \Psi_2(x) = \underbrace{A_2 e^{i k_2 x}}_{\rightarrow} + \underbrace{B_2 e^{-i k_2 x}}_{\leftarrow}, \end{align} \]

Boundary conditions:

\[ \begin{equation} \Psi_1(0) = \Psi_2(0), \end{equation} \]

\[ \begin{equation} \Psi_1'(0) = \Psi_2'(0), \end{equation} \]

(4 variables, 2 equations) - ill-defined problem?

• problem is linear, i.e. can set \(A_1=1\),
• particle incident from the left — no left-propagating component in \(i=2\):

\[ \begin{equation} \Psi_1(x) = \color{blue}{e^{i k_1 x}} + \color{orange}{B_1 e^{-i k_1 x}},\quad \Psi_2(x) = \color{green}{A_2 e^{i k_2 x}}, \end{equation} \]

…algebraic problem! \[\begin{equation} \begin{pmatrix} e^{-2ik_1x_{12}} & -e^{i(k_2-k_1)x_{12}} \\ -k_1 e^{-2ik_1x_{12}} & -k_2 e^{i(k_2-k_1)x_{12}} \end{pmatrix} \begin{pmatrix} B_1\\A_2 \end{pmatrix} = \begin{pmatrix} -1 \\ -k_1 \end{pmatrix} \end{equation} \]

solve with Wolfram Mathematica?

M = {{1,-1},
        {-1,-k2/k1}};
b = {-1, -1};
ln = LinearSolve[M,b];
B1 = ln[[1]]//FullSimplify
A2 = ln[[2]]//FullSimplify

add time evolution: \(\Psi(x,t) = e^{-iEt/\hbar}\Psi(x,0)\)

code available at github

problems:

• what’s causing the oscillating behaviour of the wavefunction in the left part?
• separate the incident and reflected waves, plot or animate,
• what happens when the energy \(E\) approaches the value of the \(V_2\)?

how to measure the transmission and reflection?

introduce current densities:

\[\begin{equation} J(x) = \frac{\hbar}{m}\Im\left\{\left[\Psi(x)\right]^* \Psi'(x)\right\}, \end{equation} \]

e.g. \[\begin{align} &J_I(x) = \frac{\hbar}{m}\Im\left\{\left[A_1 e^{ik_1 x}\right]^* A_1 i k_1 e^{ik_1 x}\right\} = \frac{\hbar k_1}{m}|A_1|^2,\\ &J_T(x) = \frac{\hbar}{m}\Im\left\{\left[A_2 e^{ik_2 x}\right]^* A_2 i k_2 e^{ik_2 x}\right\} = \frac{\hbar k_2}{m}|A_2|^2,\\ &J_R(x) = \frac{\hbar}{m}\Im\left\{\left[B_1 e^{-ik_1 x}\right]^* \left(-B_1 i k_1 e^{-ik_1 x}\right)\right\} = \frac{\hbar k_1 }{m}|B_1|^2, \end{align}\]

and transmission \(\mathcal{T}\) and reflection probabilities \(\mathcal{R}\): \[\begin{equation} \mathcal{T}= \frac{J_T}{J_I} = \frac{k_2}{k_1}\left|\frac{A_2}{A_1}\right|^2, \quad \mathcal{R}= \frac{J_R}{J_I} = \left|\frac{B_1}{A_1}\right|^2. \end{equation} \]

for \(V_1=0\) and \(V_2=1\), transmission \(\mathcal{T}\) and reflection \(\mathcal{R}\) depend on the energy \(E\) as

problems:

• what’s the limit of transmission for \(E \rightarrow V_2\)?
• what’s the significance of \(\mathcal{T}+\mathcal{R}=1\)?

what if \(V(x)>E\) somewhere?

• classical “hard border” (no penetration)

• quantum transmission

in domain 1 (\(x<0\)) — free particle \[ \begin{equation} \Bigg[\frac{\mathrm{d}^2}{\mathrm{d}x^2} + \underbrace{\frac{2m}{\hbar^2} E}_{k_1^2} \Bigg] \Psi_1(x) = 0 \rightarrow \Psi_1(x) = \underbrace{A_1 e^{i k_1 x}}_{\rightarrow} + \underbrace{B_1 e^{-i k_1 x}}_{\leftarrow}, \end{equation} \]

in domain 3 (\(x>L\)) — free particle \[ \begin{equation} \Bigg[\frac{\mathrm{d}^2}{\mathrm{d}x^2} + \underbrace{\frac{2m}{\hbar^2} E}_{k_3^2} \Bigg] \Psi_3(x) = 0 \rightarrow \Psi_3(x) = \underbrace{A_3 e^{i k_3 x}}_{\rightarrow} + \underbrace{B_3 e^{-i k_3 x}}_{\leftarrow}, \end{equation} \]

in domain 2 (\(0<x<L\)) \[ \begin{equation} \Bigg[\frac{\mathrm{d}^2}{\mathrm{d}x^2} + \underbrace{\frac{2m}{\hbar^2} (E - V_2)}_{-\kappa_2^2} \Bigg] \Psi_2(x) = 0 \rightarrow \Psi_2(x) = \underbrace{B_2 e^{-\kappa_2 x}}_{\searrow} + \underbrace{A_2 e^{\kappa_2 x}}_{\nearrow}. \end{equation} \]

\[ \begin{equation} \Psi_2(x) = \color{orange}{\underbrace{B_2 e^{-\kappa_2 x}}_{\searrow}} + \color{blue}{\underbrace{A_2 e^{\kappa_2 x}}_{\nearrow}}. \end{equation} \]

Stitching solutions

\[ \begin{align} &\Psi_1(0) = \Psi_2(0),\quad \Psi_1'(0) = \Psi_2'(0)\\ &\Psi_2(L) = \Psi_3(L),\quad \Psi_2'(L) = \Psi_3'(L) \end{align} \]

6 variables \((A_1, B_1, A_2, B_2, A_3, B_3)\) but only 4 equations! note: no left-propagating fields in domain 3 (\(B_3=0\)), linear (\(A_1=1\)):

\[ \begin{align} &\Psi_1(x) = \color{blue}{\underbrace{e^{i k_1 x}}_{\rightarrow}} + \color{orange}{\underbrace{B_1 e^{-i k_1 x}}_{\leftarrow}},\\ &\Psi_2(x) = \underbrace{B_2 e^{-\kappa_2 x}}_{\searrow} + \underbrace{A_2 e^{\kappa_2 x}}_{\nearrow},\\ &\Psi_3(x) = \color{green}{\underbrace{A_3 e^{i k_3 x}}_{\rightarrow}}, \end{align} \]

what’s the results? see you at the next lecture!